Problem 17 - Calculus

Question

Find the derivatives of the functions in \(1-26\). $$ \sec ^{2} x-\tan ^{2} x $$

Step-by-Step Solution

Answer

The derivative of the function \( \sec^2 x - \tan^2 x \) is 0.

Step 1: Identify the Functions

The given expression is \( \sec^2 x - \tan^2 x \). To find the derivative, we need to understand the individual components of the expression. These are \( \sec^2 x \) and \( \tan^2 x \).

Step 2: Differentiate \(\sec^2 x\)

The derivative of \( \sec^2 x \) can be found using the chain rule. Let \( u = \sec x \), then \( \frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx} \). Since the derivative of \( \sec x \) is \( \sec x \tan x \), we have:\[\frac{d}{dx}(\sec^2 x) = 2 \sec x \cdot \sec x \tan x = 2 \sec^2 x \tan x.\]

Step 3: Differentiate \(\tan^2 x\)

Similarly, differentiate \( \tan^2 x \) using the chain rule. Let \( v = \tan x \), so \( \frac{d}{dx}(v^2) = 2v \cdot \frac{dv}{dx} \). The derivative of \( \tan x \) is \( \sec^2 x \), so:\[\frac{d}{dx}(\tan^2 x) = 2 \tan x \cdot \sec^2 x = 2 \tan x \sec^2 x.\]

Step 4: Apply the Difference Rule

Now that we have the derivatives of both \( \sec^2 x \) and \( \tan^2 x \), we can apply the difference rule. The derivative of the difference is the difference of the derivatives:\[\frac{d}{dx}(\sec^2 x - \tan^2 x) = \frac{d}{dx}(\sec^2 x) - \frac{d}{dx}(\tan^2 x).\]

Step 5: Simplify the Derivative Expression

Substituting the derivatives from Steps 2 and 3 into our expression, we get:\[\frac{d}{dx}(\sec^2 x - \tan^2 x) = 2 \sec^2 x \tan x - 2 \tan x \sec^2 x.\]Simplifying yields:\[0.\]

Step 6: Interpret the Result

The derivative of the expression \( \sec^2 x - \tan^2 x \) is 0. This suggests that the expression is a constant with respect to \( x \). Indeed, by trigonometric identity, \( \sec^2 x - \tan^2 x = 1 \), which is constant.

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus for differentiating composite functions. Imagine you have a function within a function, like an onion with many layers. To peel off each layer, you need to understand how the derivative of one layer affects the next.

Here's the basic idea: if you have a function composed of an outer function and an inner function, the derivative of this composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In mathematical terms, if you have a function \(y = f(g(x))\), then the chain rule states:

• \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\)

In the solution, differentiating \(\sec^2 x\) involves using the chain rule. We first let \(u = \sec x\) and find \(\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx}\). Knowing that the derivative of \(\sec x\) is \(\sec x \tan x\), you can piece together the derivative of \(\sec^2 x\). The same applies to \(\tan^2 x\), where you differentiate by letting \(v = \tan x\) first. Understanding the chain rule is critical to successfully tackling more complex calculus problems.

Trigonometric Derivatives

Trigonometric derivatives are foundational to calculus, particularly when working with functions involving sine and cosine. However, other trigonometric functions have their unique derivatives that are crucial to learn.

For instance, the derivative of \(\sec x\) is \(\sec x \tan x\), and for \(\tan x\), it is \(\sec^2 x\). These derivatives arise from the fundamental trigonometric identities and interplay between trigonometric functions. Learning these derivatives is like having a toolset to solve various types of problems involving trigonometric expressions.

In our step-by-step solution, knowing the derivative of \(\sec x\) allows us to apply the chain rule correctly to find the derivative of \(\sec^2 x\). Similarly, knowing that the derivative of \(\tan x\) is \(\sec^2 x\) is vital for differentiating \(\tan^2 x\) with accuracy. These derivatives not only help in differentiating trigonometric functions but also in simplifying expressions through identity transformations.

Difference Rule

The difference rule is a straightforward principle in calculus. It states that the derivative of the difference of two functions is the difference of their derivatives. It's a simple yet powerful tool, especially when used in combination with other rules like the product and quotient rules.

In the problem at hand, we apply the difference rule to separate the derivative of \(\sec^2 x - \tan^2 x\). This involves finding the derivative of \(\sec^2 x\) and \(\tan^2 x\), and then subtracting the latter from the former. Mathematically, if \(f(x) = u(x) - v(x)\), then the rule is represented as:

• \(f'(x) = u'(x) - v'(x)\)

Applying the difference rule in this exercise highlights the beauty of calculus: seemingly complex expressions become manageable by applying fundamental rules. Ultimately, it reveals a trigonometric identity that simplifies the computation significantly, emphasizing the consistency and predictability of calculus techniques.